In five successive measurements, the mass of | vedclass

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(C) The mean mass of the ball is calculated as:$\bar{M} = \frac{2.61 + 2.58 + 2.40 + 2.73 + 2.80}{5} = \frac{13.12}{5} = 2.624 \,g \approx 2.62 \,g$.T

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$0.11$

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(C) The mean mass of the ball is calculated as:$\bar{M} = \frac{2.61 + 2.58 + 2.40 + 2.73 + 2.80}{5} = \frac{13.12}{5} = 2.624 \,g \approx 2.62 \,g$.The absolute errors in each measurement are:$|\Delta M_1| = |2.62 - 2.61| = 0.01 \,g$$|\Delta M_2| = |2.62 - 2.58| = 0.04 \,g$$|\Delta M_3| = |2.62 - 2.40| = 0.22 \,g$$|\Delta M_4| = |2.62 - 2.73| = 0.11 \,g$$|\Delta M_5| = |2.62 - 2.80| = 0.18 \,g$The mean absolute error is the average of these absolute errors:$\Delta \bar{M} = \frac{0.01 + 0.04 + 0.22 + 0.11 + 0.18}{5} = \frac{0.56}{5} = 0.112 \,g \approx 0.11 \,g$.

In five successive measurements, the mass of a ball is measured to be $2.61 \,g, 2.58 \,g, 2.40 \,g, 2.73 \,g$ and $2.80 \,g$. The mean absolute error in the measurement is (in $\,g$)

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